Let $m\ge3$ be an integer and $\zeta=\exp(-2\pi i/m)$ be the $m$-th primitive of unity. I am interested in some invariant subfields $F=K^G$ of the cyclotomic field $K=\mathbb Q(\zeta)$ corresponding to cyclic subgroups $G=\langle\sigma\rangle\le Gal(K/\mathbb Q)\cong \mathbb Z^\times$.
I found that $F=\mathbb Q(\zeta+\zeta^{-1})$ when $\sigma(\zeta)=\zeta^{-1}$.
However, for a prime number $p\in\mathbb Z_m^\times$ of order $d$ (i.e. for the smallest $d>0$ such that $p^d=1 \pmod m$), I am expecting that $F=\mathbb Q(\xi)$ for $\xi=\zeta+\zeta^p+\dots+\zeta^{p^{d-1}}$ is the invariant subfield of $K$ corresponding to $G=\langle\sigma\rangle$ for $\sigma:\zeta\mapsto\zeta^p$. It is trivial that $F\ge \mathbb Q(\xi)$, but it was difficult to show the equality.
Even I am not sure if it is a correct or not. Is there any counter example, or can we show it?
This is a proof in the case where $m$ is squarefree. For non-squarefree $m$ the linear independence claim explained below is false, but it does not immediately imply contradiction since the linear dependence relations need to be of a specific form.
Notation: I used $G$ to denote the Galois group $Gal(K/\mathbb{Q})$ and $H$ to denote the subgroup $\langle \sigma \rangle$.
Each element of the group $G = Gal(K/\mathbb{Q})$ is defined by its action on $\zeta: \tau(\zeta) = \zeta^i$ for some $i \in \mathbb{Z}_m^\times$.
When $\tau$ ranges over all elements of $G$, $\tau(\xi)$ will take the values $\zeta^i + \zeta^{ip} + \ldots + \zeta^{ip^{d-1}}$ for all $i \in \mathbb{Z}_m^\times$.
Since $\zeta^m = 1$, we can regard the exponents as being reduced modulo $m$. Since $p$ has order $d$ in $\mathbb{Z}_m^\times$, multiplication by $p$ induces a permutation on $\mathbb{Z}_m^\times$ consisting of disjoint $d$-cycles, and the number of cycles will be $k = \frac{\varphi(m)}{d}$.
Hence there will be $k$ distinct sets of exponents $\{i, ip, \ldots, ip^{d-1}\} \pmod m$ corresponding to each cycle.
If $m$ is squarefree then the set $\{\zeta^i: i \in \mathbb{Z}_m^\times\}$ is linearly independent over $\mathbb{Q}$ so each cycle corresponds to distinct values of $\tau(\xi)$. The linear independence claim has been mentioned here.
Therefore $\tau(\xi)$ takes on $k$ distinct values, which means $\xi$ has $k$ distinct Galois conjugates and hence $[\mathbb{Q}(\xi):\mathbb{Q}] = k$.
Note that if $F = K^H$ is the fixed field then by the Galois correspondence, $[K:F] = |H| = |\sigma| = d$, and therefore $[F:\mathbb{Q}] = [K:\mathbb{Q}]/[K:F] = \varphi(m)/d = k$.
Since $F \supseteq \mathbb{Q}(\xi)$ and their degrees are equal this means $F = \mathbb{Q}(\xi)$.